A solution that is formed by combining 200 ml of 0.15 M HCl with 300 ml of 0.090 M NaOH has an OH- concentration of
a. 1.0 x 10-7
b. 1.7 x 10-12
c. 0.030
d. 0.0030
e. 3.3 x 10-12

The answer is B but I do not know how to do it. Can anybody explain me how to solve this type of question? Thanks

ive been staring at this question for 5 min and dont have a clue how to do it...im curious!!! someone post already

HCl --> 0.2L*0.15M = 0.03mol
NaOH --> 0.3L*0.09M = 0.027mol

0.03-0.027 = 0.003mol of H+

0.003mol/0.5L = 0.006M of H+

pH = -log0.006M = 2.22

so, pOH = 14-2.22 = 11.78

[OH] = 10^-11.78 = 1.66*10^-12

thanks baedero

I THINK I GOT IT AFTER TRYING TO FIGURE IT OUT FOR 15 MIN!!!!

ok so you have .15M HCl x .2L = .03 moles HCL and then you have .09M NaOH x .3L = .027 moles NaOH...now I think you subtract .03-.027=.003 moles but i dont know why you would do this thats where I am getting lost..

Then my logic tells me to divide .003 by .5L of the new combined solution...this gives you .006M of the new solution

Now you use Kw=[h30][oh-] to find oh- concentration...you get 1x10^-14/.006= 1.667 x10^-12

Question: why do you subtract the moles, logically someone give me an answer...what is the reason behind subtracting .03-.027? thanks

yes, i subtract to get final pH of that solution.
remember HCl and NaOH are strong acid and base. after neutralization reaction, H+ is remained in that solution. based on that idea, you can know the pH and also pOH of that solution.

uhhh a strong acid and a strong base make a salt and water

HCl + NaOH ----> NaCl + H2O

Are you sure a strong base and a strong acid produce H+ cuz from this it doesnt seem like it....

NVM I got it, its because .03 moles of HCl is going to react with .027 moles of NaOH (limiting reactant) and the left over HCl moles in solution will be .03-.027....after neutralization of the reaction you will have .003 moles of HCl still in solution so that is where you get H+ concentration...but instead of doing it your way to find the pOH and all that you can just use the kw=1x10^-14 equation...much faster and easier to find OH- concentration

this is that one question i'm skipping on the test.

Hey guys..great question.

Instead of subtracting the moles of NaOH from the moles of HCl..why can't we subtract the Molarity of NaOH(.027 mol/.5 L=.054M) from the Molarity of HCl(.03 mol/.5L=.6M)? Can someone please explain theoretically why we can't use Molarity?

Thank u!!!

hey Tina,
Yes you can subtract the Molarity of NaOH from Molarity of HCL, but first of all Molarity of HCL would be 0.03mol/0.5L = 0.06M not 0.6M as you wrote in the question.
So, when we do the subtraction, we get 0.06-0.054=0.006M of HCL.
Now, we can use the formula
[H+]*[OH-]=Kw

0.006 * [OH-] = 1*10^(-14)
[OH-]= 1*10^(-14)/0.006
[OH-]= 1.67*10^(-12)

In this way, we don't need to do pH and pOH work and we could save our time!!

thanks sagar!